4. Clustering and classification

Exercises of week 4 are based on data from MASS library. Data Boston consists of housing Values in suburbs of Boston, and more information from the data can be found here. Data contains information for example crime rates which will be analyzed more in detail.

library(MASS) #reading the library
data("Boston") #reading data
str(Boston) # 506 obs. of  14 variables
## 'data.frame':    506 obs. of  14 variables:
##  $ crim   : num  0.00632 0.02731 0.02729 0.03237 0.06905 ...
##  $ zn     : num  18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
##  $ indus  : num  2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
##  $ chas   : int  0 0 0 0 0 0 0 0 0 0 ...
##  $ nox    : num  0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
##  $ rm     : num  6.58 6.42 7.18 7 7.15 ...
##  $ age    : num  65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
##  $ dis    : num  4.09 4.97 4.97 6.06 6.06 ...
##  $ rad    : int  1 2 2 3 3 3 5 5 5 5 ...
##  $ tax    : num  296 242 242 222 222 222 311 311 311 311 ...
##  $ ptratio: num  15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
##  $ black  : num  397 397 393 395 397 ...
##  $ lstat  : num  4.98 9.14 4.03 2.94 5.33 ...
##  $ medv   : num  24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
dim(Boston) # 506 rows and 14 columns
## [1] 506  14

Graphical overview

pairs(Boston) ## plotting matrix of the variables

From the pairs plot we can see the distributions and relationships between the variables. Because there are so many variables the picture is not so clear.

Next I will look at the correlations between variables in the data.

library(corrplot) 
## corrplot 0.84 loaded
library(tidyverse)
## -- Attaching packages --------------------------------------- tidyverse 1.3.0 --
## v ggplot2 3.3.2     v purrr   0.3.4
## v tibble  3.0.4     v dplyr   1.0.2
## v tidyr   1.1.2     v stringr 1.4.0
## v readr   1.4.0     v forcats 0.5.0
## -- Conflicts ------------------------------------------ tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()
## x dplyr::select() masks MASS::select()
cor_matrix <-cor(Boston) %>% round(digits=2) #calculating correlation matrix
cor_matrix #printing the matrix
##          crim    zn indus  chas   nox    rm   age   dis   rad   tax ptratio
## crim     1.00 -0.20  0.41 -0.06  0.42 -0.22  0.35 -0.38  0.63  0.58    0.29
## zn      -0.20  1.00 -0.53 -0.04 -0.52  0.31 -0.57  0.66 -0.31 -0.31   -0.39
## indus    0.41 -0.53  1.00  0.06  0.76 -0.39  0.64 -0.71  0.60  0.72    0.38
## chas    -0.06 -0.04  0.06  1.00  0.09  0.09  0.09 -0.10 -0.01 -0.04   -0.12
## nox      0.42 -0.52  0.76  0.09  1.00 -0.30  0.73 -0.77  0.61  0.67    0.19
## rm      -0.22  0.31 -0.39  0.09 -0.30  1.00 -0.24  0.21 -0.21 -0.29   -0.36
## age      0.35 -0.57  0.64  0.09  0.73 -0.24  1.00 -0.75  0.46  0.51    0.26
## dis     -0.38  0.66 -0.71 -0.10 -0.77  0.21 -0.75  1.00 -0.49 -0.53   -0.23
## rad      0.63 -0.31  0.60 -0.01  0.61 -0.21  0.46 -0.49  1.00  0.91    0.46
## tax      0.58 -0.31  0.72 -0.04  0.67 -0.29  0.51 -0.53  0.91  1.00    0.46
## ptratio  0.29 -0.39  0.38 -0.12  0.19 -0.36  0.26 -0.23  0.46  0.46    1.00
## black   -0.39  0.18 -0.36  0.05 -0.38  0.13 -0.27  0.29 -0.44 -0.44   -0.18
## lstat    0.46 -0.41  0.60 -0.05  0.59 -0.61  0.60 -0.50  0.49  0.54    0.37
## medv    -0.39  0.36 -0.48  0.18 -0.43  0.70 -0.38  0.25 -0.38 -0.47   -0.51
##         black lstat  medv
## crim    -0.39  0.46 -0.39
## zn       0.18 -0.41  0.36
## indus   -0.36  0.60 -0.48
## chas     0.05 -0.05  0.18
## nox     -0.38  0.59 -0.43
## rm       0.13 -0.61  0.70
## age     -0.27  0.60 -0.38
## dis      0.29 -0.50  0.25
## rad     -0.44  0.49 -0.38
## tax     -0.44  0.54 -0.47
## ptratio -0.18  0.37 -0.51
## black    1.00 -0.37  0.33
## lstat   -0.37  1.00 -0.74
## medv     0.33 -0.74  1.00

Correlations plot:

corrplot(cor_matrix, method="circle",type = "upper", cl.pos = "b", tl.pos = "d", tl.cex = 0.6 )

From the correlation plot, we can see that there are high positive and negative correlations between certain variables in the data. From the correlation matrix above we can see also the numerical correlations between variables.

summary(Boston)
##       crim                zn             indus            chas        
##  Min.   : 0.00632   Min.   :  0.00   Min.   : 0.46   Min.   :0.00000  
##  1st Qu.: 0.08205   1st Qu.:  0.00   1st Qu.: 5.19   1st Qu.:0.00000  
##  Median : 0.25651   Median :  0.00   Median : 9.69   Median :0.00000  
##  Mean   : 3.61352   Mean   : 11.36   Mean   :11.14   Mean   :0.06917  
##  3rd Qu.: 3.67708   3rd Qu.: 12.50   3rd Qu.:18.10   3rd Qu.:0.00000  
##  Max.   :88.97620   Max.   :100.00   Max.   :27.74   Max.   :1.00000  
##       nox               rm             age              dis        
##  Min.   :0.3850   Min.   :3.561   Min.   :  2.90   Min.   : 1.130  
##  1st Qu.:0.4490   1st Qu.:5.886   1st Qu.: 45.02   1st Qu.: 2.100  
##  Median :0.5380   Median :6.208   Median : 77.50   Median : 3.207  
##  Mean   :0.5547   Mean   :6.285   Mean   : 68.57   Mean   : 3.795  
##  3rd Qu.:0.6240   3rd Qu.:6.623   3rd Qu.: 94.08   3rd Qu.: 5.188  
##  Max.   :0.8710   Max.   :8.780   Max.   :100.00   Max.   :12.127  
##       rad              tax           ptratio          black       
##  Min.   : 1.000   Min.   :187.0   Min.   :12.60   Min.   :  0.32  
##  1st Qu.: 4.000   1st Qu.:279.0   1st Qu.:17.40   1st Qu.:375.38  
##  Median : 5.000   Median :330.0   Median :19.05   Median :391.44  
##  Mean   : 9.549   Mean   :408.2   Mean   :18.46   Mean   :356.67  
##  3rd Qu.:24.000   3rd Qu.:666.0   3rd Qu.:20.20   3rd Qu.:396.23  
##  Max.   :24.000   Max.   :711.0   Max.   :22.00   Max.   :396.90  
##      lstat            medv      
##  Min.   : 1.73   Min.   : 5.00  
##  1st Qu.: 6.95   1st Qu.:17.02  
##  Median :11.36   Median :21.20  
##  Mean   :12.65   Mean   :22.53  
##  3rd Qu.:16.95   3rd Qu.:25.00  
##  Max.   :37.97   Max.   :50.00

From the summary table we can see the mean and median values of variables. Also min and max values and quartiles are visible.

Standardize the dataset and print out summaries of the scaled data. How did the variables change? Create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate). Use the quartiles as the break points in the categorical variable. Drop the old crime rate variable from the dataset. Divide the dataset to train and test sets, so that 80% of the data belongs to the train set. (0-2 points)

Scaling the data

Next I will center and standardize the variables by function scale

boston_scaled <- scale(Boston)
summary(boston_scaled)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865

If we now look at the summary table, we can see that the values have changed. All of the mean values are 0, and almost all of the median, 1st Qu, and min values are negative.

boston_scaled <- as.data.frame(boston_scaled) #changing the object to data frame
class(boston_scaled)
## [1] "data.frame"

Next I will create a categorical variable of the crime rate in the Boston dataset (from the scaled crime rate).

summary(boston_scaled$crim)
##      Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
## -0.419367 -0.410563 -0.390280  0.000000  0.007389  9.924110
bins <- quantile(boston_scaled$crim) #creating a quantile vector of crim and printing it
bins
##           0%          25%          50%          75%         100% 
## -0.419366929 -0.410563278 -0.390280295  0.007389247  9.924109610
# creating a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label=c("low", "med_low", "med_high", "high"))
table(crime)
## crime
##      low  med_low med_high     high 
##      127      126      126      127
boston_scaled <- dplyr::select(boston_scaled, -crim)# removing original crim from the dataset

boston_scaled <- data.frame(boston_scaled, crime)# adding the new categorical value to scaled data

str(boston_scaled) #looking that everything worked
## 'data.frame':    506 obs. of  14 variables:
##  $ zn     : num  0.285 -0.487 -0.487 -0.487 -0.487 ...
##  $ indus  : num  -1.287 -0.593 -0.593 -1.306 -1.306 ...
##  $ chas   : num  -0.272 -0.272 -0.272 -0.272 -0.272 ...
##  $ nox    : num  -0.144 -0.74 -0.74 -0.834 -0.834 ...
##  $ rm     : num  0.413 0.194 1.281 1.015 1.227 ...
##  $ age    : num  -0.12 0.367 -0.266 -0.809 -0.511 ...
##  $ dis    : num  0.14 0.557 0.557 1.077 1.077 ...
##  $ rad    : num  -0.982 -0.867 -0.867 -0.752 -0.752 ...
##  $ tax    : num  -0.666 -0.986 -0.986 -1.105 -1.105 ...
##  $ ptratio: num  -1.458 -0.303 -0.303 0.113 0.113 ...
##  $ black  : num  0.441 0.441 0.396 0.416 0.441 ...
##  $ lstat  : num  -1.074 -0.492 -1.208 -1.36 -1.025 ...
##  $ medv   : num  0.16 -0.101 1.323 1.182 1.486 ...
##  $ crime  : Factor w/ 4 levels "low","med_low",..: 1 1 1 1 1 1 2 2 2 2 ...

Next I will divide the dataset to train and test sets, so that 80% of the data belongs to the train set.

n <- nrow(boston_scaled)# number of rows in the Boston dataset 

ind <- sample(n,  size = n * 0.8)# choosing randomly 80% of the rows

train <- boston_scaled[ind,] # creating train set containing 80 % of the data

test <- boston_scaled[-ind,] # creating test set containing 20 % of the data

Linear discriminant analysis

Next I will fit the linear discriminant analysis on the train set. I will use the categorical crime rate as the target variable and all the other variables in the dataset as predictor variables.

lda.fit <- lda(crime ~., data = train) # linear discriminant analysis
lda.fit # print the lda.fit object
## Call:
## lda(crime ~ ., data = train)
## 
## Prior probabilities of groups:
##       low   med_low  med_high      high 
## 0.2400990 0.2574257 0.2500000 0.2524752 
## 
## Group means:
##                  zn      indus         chas        nox          rm        age
## low       0.9349989 -0.9340350 -0.109974419 -0.8717619  0.40288933 -0.8817135
## med_low  -0.1058810 -0.2417528 -0.007331936 -0.5651173 -0.12707950 -0.3475990
## med_high -0.4023350  0.2134165  0.195445218  0.4661624 -0.04641872  0.4145341
## high     -0.4872402  1.0171096 -0.002135914  1.0764482 -0.34114341  0.7904963
##                 dis        rad        tax     ptratio       black      lstat
## low       0.8560124 -0.6764026 -0.7483441 -0.40327083  0.38385860 -0.7722106
## med_low   0.3494809 -0.5500918 -0.4598202 -0.04652584  0.30754012 -0.1566894
## med_high -0.4157672 -0.4190087 -0.3097095 -0.29959318  0.05864284  0.1069822
## high     -0.8393948  1.6382099  1.5141140  0.78087177 -0.68991502  0.8941351
##                 medv
## low       0.51418896
## med_low  -0.01328999
## med_high  0.04315452
## high     -0.64507330
## 
## Coefficients of linear discriminants:
##                 LD1         LD2         LD3
## zn       0.09122829  0.66665268 -0.96202543
## indus    0.05802113 -0.23892754  0.53596350
## chas    -0.07809839 -0.05534393  0.11547559
## nox      0.33367634 -0.70242001 -1.39041153
## rm      -0.10388361 -0.05848535 -0.02553381
## age      0.22078324 -0.33149639 -0.07032695
## dis     -0.09344412 -0.20024390  0.35607716
## rad      3.22338975  0.87579229 -0.14157562
## tax     -0.06982094  0.06840261  0.64996947
## ptratio  0.11159391  0.05749545 -0.30643711
## black   -0.10036931  0.01368226  0.09409443
## lstat    0.21216968 -0.17785315  0.24532524
## medv     0.19201862 -0.26006836 -0.26695304
## 
## Proportion of trace:
##    LD1    LD2    LD3 
## 0.9460 0.0397 0.0143

I will draw the LDA (bi)plot, but first we need the function for lda biplot arrows.

lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
  heads <- coef(x)
  arrows(x0 = 0, y0 = 0, 
         x1 = myscale * heads[,choices[1]], 
         y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
  text(myscale * heads[,choices], labels = row.names(heads), 
       cex = tex, col=color, pos=3)
}

classes <- as.numeric(train$crime) # target classes as numeric
plot(lda.fit, dimen = 2, col = classes, pch = classes)# plotting the lda results
lda.arrows(lda.fit, myscale = 2) #adding the arrows

Plot looks different than the one that I did in the DataCamp exercise, but because R takes randomly 80 % of the Boston data to the train data, this 80% might be now different than in DataCamp.

Model predictions

Next I will save the crime categories from the test set and then remove the categorical crime variable from the test dataset. Then I will predict the classes with the LDA model on the test data.

correct_classes <- test$crime # saving the correct classes from test data

test <- dplyr::select(test, -crime) # removing the crime variable from test data
lda.pred <- predict(lda.fit, newdata = test) # predicting classes with test data

table(correct = correct_classes, predicted = lda.pred$class) # cross tabulating the results
##           predicted
## correct    low med_low med_high high
##   low       17      10        3    0
##   med_low    3      14        5    0
##   med_high   0       6       18    1
##   high       0       0        0   25

If we look at the prediction results, we can see that it works quite well. Especially in high there is only 1 observation which is predicted as med_high. In other classes about 50 % of the predicted observations are correct.

K-means clustering

Reload the Boston dataset and standardize the dataset (we did not do this in the Datacamp exercises, but you should scale the variables to get comparable distances).

data(Boston)#reloading the Boston dataset
boston_scaled <- scale(Boston) #scaling
summary(boston_scaled)
##       crim                 zn               indus              chas        
##  Min.   :-0.419367   Min.   :-0.48724   Min.   :-1.5563   Min.   :-0.2723  
##  1st Qu.:-0.410563   1st Qu.:-0.48724   1st Qu.:-0.8668   1st Qu.:-0.2723  
##  Median :-0.390280   Median :-0.48724   Median :-0.2109   Median :-0.2723  
##  Mean   : 0.000000   Mean   : 0.00000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.007389   3rd Qu.: 0.04872   3rd Qu.: 1.0150   3rd Qu.:-0.2723  
##  Max.   : 9.924110   Max.   : 3.80047   Max.   : 2.4202   Max.   : 3.6648  
##       nox                rm               age               dis         
##  Min.   :-1.4644   Min.   :-3.8764   Min.   :-2.3331   Min.   :-1.2658  
##  1st Qu.:-0.9121   1st Qu.:-0.5681   1st Qu.:-0.8366   1st Qu.:-0.8049  
##  Median :-0.1441   Median :-0.1084   Median : 0.3171   Median :-0.2790  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.5981   3rd Qu.: 0.4823   3rd Qu.: 0.9059   3rd Qu.: 0.6617  
##  Max.   : 2.7296   Max.   : 3.5515   Max.   : 1.1164   Max.   : 3.9566  
##       rad               tax             ptratio            black        
##  Min.   :-0.9819   Min.   :-1.3127   Min.   :-2.7047   Min.   :-3.9033  
##  1st Qu.:-0.6373   1st Qu.:-0.7668   1st Qu.:-0.4876   1st Qu.: 0.2049  
##  Median :-0.5225   Median :-0.4642   Median : 0.2746   Median : 0.3808  
##  Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 1.6596   3rd Qu.: 1.5294   3rd Qu.: 0.8058   3rd Qu.: 0.4332  
##  Max.   : 1.6596   Max.   : 1.7964   Max.   : 1.6372   Max.   : 0.4406  
##      lstat              medv        
##  Min.   :-1.5296   Min.   :-1.9063  
##  1st Qu.:-0.7986   1st Qu.:-0.5989  
##  Median :-0.1811   Median :-0.1449  
##  Mean   : 0.0000   Mean   : 0.0000  
##  3rd Qu.: 0.6024   3rd Qu.: 0.2683  
##  Max.   : 3.5453   Max.   : 2.9865
boston_scaled <- as.data.frame(boston_scaled) #changing the object to data frame
class(boston_scaled)
## [1] "data.frame"

Next I will calculate the distances between the observations and run k-means algorithm on the dataset.

dist_eu <- dist(boston_scaled) # euclidean distance matrix

summary(dist_eu) #summary of the distances
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.1343  3.4625  4.8241  4.9111  6.1863 14.3970
dist_man <- dist(boston_scaled, method = 'manhattan')# manhattan distance matrix

summary(dist_man)#summary of the distances
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.2662  8.4832 12.6090 13.5488 17.7568 48.8618
km <-kmeans(boston_scaled, centers = 3) # k-means clustering

pairs(boston_scaled[1:6], col = km$cluster)# plotting the scaled Boston dataset with clusters 

pairs(boston_scaled[7:14], col = km$cluster)# perform plotting in two parts so that plots are easier to look at

Next I will investigate what is the optimal number of clusters.

library(ggplot2)
set.seed(123)
k_max <- 15 #determining the number of clusters
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss}) # calculate the total within sum of squares

# visualize the results
qplot(x = 1:k_max, y = twcss, geom = 'line')

If we look at the twcss-plot, there is no huge drop in total wcss, but it looks like 5 clusters might be optimal.

I will now do the clustering again

km <-kmeans(boston_scaled, centers = 5) # now 5 centers

pairs(boston_scaled[1:6], col = km$cluster)# plotting the scaled Boston dataset with clusters 

pairs(boston_scaled[7:14], col = km$cluster)# perform plotting in two parts so that plots are easier to look at

From the plots above, we can see the clustering in different colours. Some clusters are very clearly separated, but in other cases clusters are on top on each other. So I am not quite sure what would be the most optimal number of clusters in this case.

Super bonus exercise

Super-Bonus: Run the code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.

model_predictors <- dplyr::select(train, -crime)

dim(model_predictors)# check the dimensions
## [1] 404  13
dim(lda.fit$scaling)
## [1] 13  3
# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)

I installed package: install.packages(“plotly”)

library(plotly)#accessing library
## 
## Attaching package: 'plotly'
## The following object is masked from 'package:ggplot2':
## 
##     last_plot
## The following object is masked from 'package:MASS':
## 
##     select
## The following object is masked from 'package:stats':
## 
##     filter
## The following object is masked from 'package:graphics':
## 
##     layout

Let´s create a 3D plot of the columns of the matrix product

plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers')
## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.

Adjust the code: add argument color as a argument in the plot_ly() function. Set the color to be the crime classes of the train set. Draw another 3D plot where the color is defined by the clusters of the k-means. How do the plots differ? Are there any similarities? (0-3 points to compensate any loss of points from the above exercises)